Timeline for Are there any known criteria for quadratic mapping from R^n to R^n being surjective?
Current License: CC BY-SA 3.0
9 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jan 13, 2014 at 18:37 | comment | added | David E Speyer | The paper you quote has what I consider a misleading framing of the problem: Their map contains a computable number $a$ and the map will be surjective iff $a \neq 0$; there is no algorithm which takes a description of a computable number and determines whether it is zero. If you give your mapping $f$ in a more concrete manner (for example, with coefficients in $\mathbb{Q}$), then this will be computable by Tarski's Theorem mathworld.wolfram.com/TarskisTheorem.html . | |
| Jan 4, 2014 at 12:11 | comment | added | Alexander Chervov | Disagree with J. Martel, at least for me, joro's answer nice "sharing of knowledge" even if it does not precisely match the question | |
| Jan 2, 2014 at 20:45 | comment | added | JHM | @joro: in my opinion your `answer' needlessly clutters this particular question and deserves to be deleted. | |
| Dec 31, 2013 at 16:13 | history | edited | joro | CC BY-SA 3.0 |
Rollback, misread the question
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| Dec 31, 2013 at 16:11 | comment | added | joro | @PietroMajer thanks, I see, misread the question. | |
| Dec 31, 2013 at 16:05 | comment | added | Pietro Majer | 1) Here the polynomial mapping is polynomial, but of quite a special form. Is the quoted proposition true even in this very particular class? 2) The OP means a map whose coordinates are quadratic forms, i.e. homogeneous polynomials of degree 2. If we allow linear terms then it is very easy to build examples of invertible maps. | |
| Dec 31, 2013 at 15:59 | history | edited | joro | CC BY-SA 3.0 |
Added example per discussion with abx
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| Dec 31, 2013 at 13:25 | history | answered | joro | CC BY-SA 3.0 |